Find the exact value of #sin(pi/12)# and #cos(pi/12)#?

Question

2484 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-01T17:26:53

#sin(pi/12)=(sqrt3-1)/(2sqrt2)#

#cos(pi/12)=(sqrt3+1)/(2sqrt2)#

As #cos2A=2cos^2A-1#

#cos(2xxpi/12)=2cos^2(pi/12)-1#

or #cos(pi/6)=sqrt3/2=2cos^2(pi/12)-1#

or #2cos^2(pi/12)-1-sqrt3/2=0#

or #2cos^2(pi/12)-(2+sqrt3)/2=0#

or #cos^2(pi/12)=(2+sqrt3)/4#

and #cos(pi/12)=sqrt(2+sqrt3)/2=sqrt(4+2sqrt3)/(2sqrt2)#

= #sqrt(3+1+2sqrt3)/(2sqrt2)#

= #sqrt((sqrt3)^2+1^2+2xxsqrt3xx1)/(2sqrt2)#

= #(sqrt3+1)/(2sqrt2)#

Further #cos2A=1-2sin^2A#

#cos(pi/6)=1-2sin^2(pi/12)#

or #sqrt3/2=1-2sin^2(pi/12)#

or #2sin^2(pi/12)=1-sqrt3/2#

or #sin^2(pi/12)=(2-sqrt3)/4#

or #sin(pi/12)=(sqrt(2-sqrt3)/2#

= #sqrt(4-2sqrt3)/(2sqrt2)#

= #sqrt(3+1-2sqrt3)/(2sqrt2)#

= #sqrt((sqrt3)^2+1^2-2xxsqrt3xx1)/(2sqrt2)#

= #(sqrt3-1)/(2sqrt2)#